Jump Game Problem in DSA C++ - Time & Space Complexity

Time Complexity: Jump Game Problem

O(n)

Understanding Time Complexity

We want to understand how the time needed to solve the Jump Game problem changes as the input size grows.

Specifically, how does the number of steps to check jumps grow with the length of the array?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

bool canJump(std::vector<int>& nums) {
    int reachable = 0;
    for (int i = 0; i < nums.size(); i++) {
        if (i > reachable) return false;
        reachable = std::max(reachable, i + nums[i]);
    }
    return true;
}
    

This code checks if you can jump from the start to the end of the array, where each number tells the max jump length from that position.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop that goes through each element of the array once.
• How many times: Exactly once for each element, so n times if the array length is n.

How Execution Grows With Input

As the array gets longer, the code checks each position once to update the farthest reachable index.

Input Size (n)	Approx. Operations
10	About 10 checks
100	About 100 checks
1000	About 1000 checks

Pattern observation: The number of operations grows directly with the input size, so doubling the input roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to decide if you can reach the end grows linearly with the size of the input array.

Common Mistake

[X] Wrong: "Because we check jumps from each position, the time is quadratic O(n²)."

[OK] Correct: The code only loops once through the array and updates the farthest reachable index without nested loops, so it does not repeat work for each jump.

Interview Connect

Understanding this linear time solution shows you can analyze and optimize problems that look like they might need more work, which is a great skill for interviews.

Self-Check

"What if we changed the code to try every possible jump from each position instead of just tracking the farthest reachable? How would the time complexity change?"

Practice

1. What is the main goal of the Jump Game problem?

easy

A. Count the total number of jumps needed to reach the end

B. Find the maximum jump length from the first index

C. Calculate the minimum jump length required to reach the last index

D. Determine if you can reach the last index from the first index by jumping

Solution

1. Step 1: Understand the problem statement

   The Jump Game problem asks if it is possible to move from the first position of an array to the last position by jumping forward. Each array element represents the maximum jump length from that position.

2. Step 2: Identify the goal

   The goal is to check if the last index can be reached starting from the first index using the allowed jumps.

3. Final Answer:

   Determine if you can reach the last index from the first index by jumping -> Option D

4. Quick Check:

   Jump Game goal = Reach last index [OK]

Hint: Focus on reachability from start to end index [OK]

Common Mistakes:

• Confusing maximum jump length with total jumps
• Trying to count jumps instead of checking reachability
• Assuming jumps can be backward

2. Which of the following is the correct C++ syntax to declare a vector named nums initialized with values {2,3,1,1,4}?

easy

A. vector nums = [2,3,1,1,4];

B. std::vector nums = {2,3,1,1,4};

C. std::vector<int> nums = {2,3,1,1,4};

D. std::vector<int> nums(2,3,1,1,4);

Solution

1. Step 1: Understand vector declaration syntax

   In C++, to declare a vector of integers with initial values, use std::vector<int> nums = {values};.

2. Step 2: Check each option

   std::vector<int> nums = {2,3,1,1,4}; uses correct syntax with type and initializer list. Options A and C miss template type or use wrong brackets. std::vector<int> nums(2,3,1,1,4); uses constructor incorrectly.

3. Final Answer:

   std::vector<int> nums = {2,3,1,1,4}; -> Option C

4. Quick Check:

   Correct vector initialization syntax = std::vector<int> nums = {2,3,1,1,4}; [OK]

Hint: Use std::vector<int> with curly braces for initialization [OK]

Common Mistakes:

• Omitting <int> in vector declaration
• Using square brackets [] instead of curly braces {}
• Incorrect constructor parameters

3. What will be the output of the following C++ code snippet?

std::vector<int> nums = {2,3,1,1,4};
int maxReach = 0;
for (int i = 0; i <= maxReach; ++i) {
    maxReach = std::max(maxReach, i + nums[i]);
    if (maxReach >= nums.size() - 1) {
        std::cout << "true" << std::endl;
        break;
    }
}
if (maxReach < nums.size() - 1) std::cout << "false" << std::endl;

medium

A. false

B. true

C. Compilation error

D. Runtime error

Solution

1. Step 1: Trace the loop and maxReach updates

   Start with maxReach=0. At i=0, maxReach = max(0, 0+2) = 2. At i=1, maxReach = max(2, 1+3) = 4. Since maxReach (4) >= last index (4), print "true" and break.

2. Step 2: Confirm output

   The loop breaks after printing "true". No further output.

3. Final Answer:

   true -> Option B

4. Quick Check:

   maxReach reaches last index, output = true [OK]

Hint: Track maxReach and break when it reaches last index [OK]

Common Mistakes:

• Not updating maxReach correctly
• Missing break causing extra output
• Confusing index bounds in loop

4. Identify the error in the following C++ code for the Jump Game problem:

bool canJump(std::vector<int>& nums) {
    int maxReach = 0;
    for (int i = 0; i < nums.size(); ++i) {
        if (i > maxReach) return false;
        maxReach = std::max(maxReach, i + nums[i]);
    }
    return true;
}

medium

A. No error, code works correctly

B. Return false should be after the loop, not inside

C. maxReach update is incorrect, should subtract 1

D. Loop condition should be i <= maxReach instead of i < nums.size()

Solution

1. Step 1: Analyze loop and conditions

   The loop iterates over all indices. If current index i is beyond maxReach, return false immediately. Otherwise, update maxReach.

2. Step 2: Verify correctness

   This logic correctly checks if the last index is reachable. Returning false inside loop is valid. maxReach update is correct.

3. Final Answer:

   No error, code works correctly -> Option A

4. Quick Check:

   Code logic matches Jump Game solution [OK]

Hint: Check if current index exceeds maxReach to return false early [OK]

Common Mistakes:

• Changing loop condition incorrectly
• Moving return false outside loop causing wrong result
• Incorrect maxReach calculation

5. Given the array nums = {3,2,1,0,4}, why does the standard greedy Jump Game algorithm return false?

hard

A. Because the zero at index 3 blocks any further jumps to the last index

B. Because the array length is less than 5

C. Because the first element is too large causing an overflow

D. Because the algorithm only works for arrays with all positive numbers

Solution

1. Step 1: Understand the array and jumps

   Starting at index 0 with jump length 3, you can reach indices 1, 2, or 3. At index 3, jump length is 0, so you cannot move forward from there.

2. Step 2: Identify why last index is unreachable

   Since index 4 is beyond index 3 and no jumps can be made from index 3, the last index cannot be reached, so the algorithm returns false.

3. Final Answer:

   Because the zero at index 3 blocks any further jumps to the last index -> Option A

4. Quick Check:

   Zero jump blocks path, output = false [OK]

Hint: Look for zero jump spots blocking progress [OK]

Common Mistakes:

• Ignoring zero jump blocking
• Assuming array length affects result
• Misunderstanding jump lengths